Question

A train 400 m long overtook a man walking along the line in the same direction as the train, at the rate of 5 kmph and passed him in 40 seconds. The train reached the station in 20 minutes after passing the man. In what time did the man reach the station?

• A. 2 hr 24 min 48 sec
• B. 2 hr 24 min 24 sec
• C. 2 hr 30 min 40 sec
• D. 2 hr 36 min 48 sec
• E. 2 hr 48 min 48 sec

Hint:

Always be consistent with units (km/h or m/s). A common source of error is mixing units in the same calculation. In this case, the answer might rely on a subtlety in the problem statement.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves relative speed between the train and the man, and calculating time for a fixed distance.
Step 2: Key Formula or Approach:
Relative Speed \( = \text{Length} / \text{Time} \). Distance \( = \text{Speed} \cdot \text{Time} \).
Step 3: Detailed Explanation:
1. Find train speed:
Man's speed \( = 5 \cdot (5/18) = 25/18 \) m/s.
Relative speed \( = 400 / 40 = 10 \) m/s.
In same direction: \( V_t - V_m = 10 \Rightarrow V_t = 10 + 25/18 = 205/18 \) m/s.
Train speed in kmph \( = (205/18) \cdot (18/5) = 41 \) kmph.
2. Find distance to station:
Train head reaches station in 20 min (\( 1/3 \) hr).
Distance covered by head \( = 41 \cdot (1/3) = 41/3 \) km.
3. Total distance for man:
When train just finished passing, the man was at the train's tail.
So man is \( 400 \)m (\( 0.4 \) km) behind the current position of the train head.
Total distance for man \( = 41/3 + 0.4 = (41 + 1.2)/3 = 42.2/3 \) km.
4. Time for man \( = (42.2/3) / 5 = 42.2 / 15 \) hours.
\( 42.2 / 15 = 2.8133... \) hours.
\( 2 \text{ hr} + (0.8133 \cdot 60) \text{ min} = 48.8 \) min.
\( 48 \text{ min} + (0.8 \cdot 60) \text{ sec} = 48 \) sec.
Step 4: Final Answer:
The man reached the station in 2 hr 48 min 48 sec.